Calculating Electron Interference Pattern Minima with Double-Slit Apparatus

[erok81]

Homework Statement

Electrons are accelerated through a 20 V potential difference, producing a mono energetic beam. This is directed at a double-slit apparatus of 0.010mm slit separation. A bank of electron detectors is 10m beyond the double slit. With slit 1 alone open, 100 electrons per second are detected at all detectors. With slit 2 alone open, 900 electron per second are detected at all detectors. Now both slits are open.

The first minimum in the electron count occurs at detector X. How far is it from the center of the interference pattern?

Homework Equations

dsinθ=mλ

λ=h/p

The Attempt at a Solution

First we will need to find λ. Using the de Broglie wavelength can calculate this. However, I am stuck here.

Electron volts were introduced this semester without any direction so I don't really know how to use them. I know that an electron volt is the kinetic energy obtained by acclerating an electron over a 1V potential. So I am assuming that to find the KE in this case I can multiply 20 by 1.60x10_-19 and I will have KE in eV. Now first sticking point - how is this translated to momentum? I know the mass and KE...so I would need a velocity?

Next I am not sure how covering and uncovering affects the calculations. I understand how it affects the pattern, but that's about it. Although from my optics class do I need to know this? I can solve for theta which will give me my distance from the center...

So summarized. I need help converting my calculated eV to momentum - if I even calculated eV correctly and also how covering and uncovering the slits plays into solving this problem.

 

[lswtech]

"multiply 20 by 1.60x10-19" gives you KE in J
by 1/2 mv^2 you could get v and hence mv, neglecting relativistic effects

I thought covering one silt is "diffraction" and it follows "dsinθ=mλ when destructive", in contrast to "dsinθ=mλ when constructive" in interference

 

[erok81]

"multiply 20 by 1.60x10-19" gives you KE in J
by 1/2 mv^2 you could get v and hence mv, neglecting relativistic effects

I thought covering one silt is "diffraction" and it follows "dsinθ=mλ when destructive", in contrast to "dsinθ=mλ when constructive" in interference

Oh yeah duh. E=qV. That will give me Joules. Then I can find the velocity/momentum with E=mc²(γ-1)

From my book when convering one slit, at least with electron, the interference pattern goes away completely. So those two pieces of info needs to be used in order to finish the rest of the problem. Hmm...

 

[erok81]

Ok...wavelength complete. λ=2.746x10^-10m.

 

[Redbelly98]

I agree with your wavelength calculation. BTW, a nonrelativistic calculation would have worked too, since 20 eV is a lot less than the electron rest-mass energy of 511,000 eV.

I wouldn't worry about the fact that only one slit at a time was open. They are asking for the minimum with both slits open. Do you know the condition for destructive interference? The dsinθ=mλ equation is for constructive interference, since it essentially says the path difference is an integer multiple of the wavelength.

 

[erok81]

Ah okay, you are right. The open/closed doesn't matter until part a and b. I'll give that a try and see how it goes.

I found θ using sinθ=1/2d*λ. Which gave me an angle of 7.84e-4. So the answer for where the first minimum occurs is 1.37e-4.

Thanks for adding that part in about non-relativistic vs non. I am doing as many problems as I can find involving eV since I am not used to working with them. Your comment helped a lot in that aspect. I was always using E=mc2(γ-1) just incase it ended up being relativistic. 1/2*mv² would be a lot easier to calculate. :)

The next part is, how many electrons per second will be detected at the center detector?

Here is how I solved it. Does this look legit? I got the correct answer, but I want to make sure my method is correct.

Slit one open only.
|Ψ₁|²| ∝ 100^-1s = 10 electrons/second

Slit two open only.
|Ψ₂|²| ∝ 900^-1s = 30 electrons/second

Both slits now open. Adding both slits = 40 electrons per second.

Working backwards from above. Now Ψ = 40.

|Ψ(40)|² = 1600

I will admit, I looked at the solution and worked backwards to obtain that answer. Since I knew that the square of the amplitude of the matter wave is proportional to the particle probability detection.

 

[erok81]

Last part is c) how many electrons per second will be detected at detector X?

Detector X is the first minimum.

This one isn't so easy. I can do the same as above but instead of adding the two, I can subtract and then square getting the same answer.

So a max they are added and a min they are subtracted? Back to my text to find out why.

 

[Redbelly98]

Ah okay, you are right. The open/closed doesn't matter until part a and b. I'll give that a try and see how it goes.

I found θ using sinθ=1/2d*λ. Which gave me an angle of 7.84e-4. So the answer for where the first minimum occurs is 1.37e-4.

Agreed.

The next part is, how many electrons per second will be detected at the center detector?

Here is how I solved it. Does this look legit? I got the correct answer, but I want to make sure my method is correct.

Slit one open only.
|Ψ₁|²| ∝ 100^-1s = 10 electrons/second

Slit two open only.
|Ψ₂|²| ∝ 900^-1s = 30 electrons/second

Both slits now open. Adding both slits = 40 electrons per second.

Working backwards from above. Now Ψ = 40.

|Ψ(40)|² = 1600

Looks more or less okay, but there is some sloppiness in the equations -- I don't know if that's because you were writing in a hurry or don't understand it.

For example, the equation

|Ψ₁|²| ∝ 100^-1s = 10 electrons/second
​

Is saying that |Ψ₁|² ∝ 10, when it should say that |Ψ₁|∝10. Also, 100^-1s should be 100s^-1.

Last part is c) how many electrons per second will be detected at detector X?

Detector X is the first minimum.

This one isn't so easy. I can do the same as above but instead of adding the two, I can subtract and then square getting the same answer.

So a max they are added and a min they are subtracted? Back to my text to find out why.

In destructive interference, the two waves are 180° out of phase, so you subtract them. This produces a minimum in |Ψ|².

 

[erok81]

Agreed.

Looks more or less okay, but there is some sloppiness in the equations -- I don't know if that's because you were writing in a hurry or don't understand it.

For example, the equation

|Ψ₁|²| ∝ 100^-1s = 10 electrons/second
​

Is saying that |Ψ₁|² ∝ 10, when it should say that |Ψ₁|∝10. Also, 100^-1s should be 100s^-1.

A little of both. I typed the first one wrong and the copy/pasted it so they both came out wrong. And I see what you mean with the other mistakes. I knew the order they should go, but not how to type them out.

In destructive interference, the two waves are 180° out of phase, so you subtract them. This produces a minimum in |Ψ|².

Ah got it. That makes sense. It now makes sense on both counts. I keep forgetting these things act as waves!

Thanks again for your detailed responses. Now I actually understand the concept behind this question.

 

[Redbelly98]

Now I actually understand the concept behind this question.

Cool! Good luck.